# If the mol ratio of nitrogen and oxygen is 4:1, what is the ratio of their solubilities in terms of mol fractions? Take k_H = 3.3 xx 10^7 "torr" for "N"_2(g) and 6.6 xx 10^7 "torr" for "O"_2(g).

May 4, 2017

$2 : 1$, ${\text{N"_2:"O}}_{2}$

You should consider whether the question wants ${\text{N"_2:"O}}_{2}$ or ${\text{O"_2:"N}}_{2}$.

There are two things you should be thinking about here:

• What is Henry's law in terms of mol fractions?
• What does knowing the ratio of ${\text{N"_2:"O}}_{2}$ in air tell you about their partial pressures?

Setting up the solution is usually the hard part. Imagine a beaker of water that has air dissolved in it, but there is, say, parafilm on it so that it is a closed container. Kind of like this, but with transparent water: What you are looking for is the ratio of the solubilities on the mol fraction scale. It may or may not be $4 : 1$ or $1 : 4$.

Henry's law in terms of mol fractions for the solute in particularly-dilute solutions is:

$\boldsymbol{{P}_{j} = {\chi}_{j \left(v\right)} {k}_{H , j}}$

where:

• ${P}_{j}$ is the partial pressure of solute $j$ above the solution.
• ${\chi}_{j \left(v\right)}$ is the mol fraction of solute $j$ in the vapor phase ($v$), that is, above the solution.
• ${k}_{H , j}$ is the Henry's law constant of solute $j$, and ${k}_{H , j}$ approaches the pure vapor pressure ${P}_{j}^{\text{*}}$ as ${\chi}_{j \left(l\right)} \to 1$ (since the solute particles surround the solvent particles completely, and vaporize most easily).

Writing Henry's law for ${\text{N}}_{2}$ and ${\text{O}}_{2}$:

${P}_{{N}_{2}} = {\chi}_{{N}_{2} \left(v\right)} {k}_{H , {N}_{2}}$

${P}_{{O}_{2}} = {\chi}_{{O}_{2} \left(v\right)} {k}_{H , {O}_{2}}$

But from Dalton's law of partial pressures for ideal gases, we have:

${P}_{t o t} = {P}_{{N}_{2}} + {P}_{{O}_{2}}$

$= 0.78 {P}_{t o t} + 0.21 {P}_{t o t} + \stackrel{\text{Other}}{\overbrace{0.01 {P}_{t o t}}}$

Thus, we can rewrite Henry's law as:

$0.78 {P}_{t o t} = {\chi}_{{N}_{2} \left(v\right)} {k}_{H , {N}_{2}}$

$0.21 {P}_{t o t} = {\chi}_{{O}_{2} \left(v\right)} {k}_{H , {O}_{2}}$

Finding the ratio of their mol fraction solubilities, we then have:

$\frac{0.78 {P}_{t o t}}{0.21 {P}_{t o t}} = \frac{{\chi}_{{N}_{2} \left(v\right)} {k}_{H , {N}_{2}}}{{\chi}_{{O}_{2} \left(v\right)} {k}_{H , {O}_{2}}}$

$\implies \frac{{\chi}_{{N}_{2} \left(v\right)}}{{\chi}_{{O}_{2} \left(v\right)}} = 3.71 \left({k}_{H , {O}_{2}} / {k}_{H , {N}_{2}}\right)$

$= 3.71 \left(\left(3.3 \times {10}^{7} \text{torr")/(6.6 xx 10^7 "torr}\right)\right)$

$= 1.86$

That's using a mol ratio of $0.78 : 0.21$, as it is in reality, but using a mol ratio of $4 : 1$ like given in the problem, we would have:

$\textcolor{b l u e}{\frac{{\chi}_{{N}_{2} \left(v\right)}}{{\chi}_{{O}_{2} \left(v\right)}}} = 4 \left(\left(3.3 \times {10}^{7} \text{torr")/(6.6 xx 10^7 "torr}\right)\right)$

$= \textcolor{b l u e}{2 : 1}$, $\textcolor{b l u e}{{\text{N"_2:"O}}_{2}}$

If the question wants the ratio of ${\text{O}}_{2}$ to ${\text{N}}_{2}$, then in that case it would be $1 : 2$, ${\text{O"_2:"N}}_{2}$.

In the end, it is just the mol ratio in air, weighted by the reciprocal ratio of the Henry's law constants.